AQA A-Level Chemistry Paper 3, 2023: Question 17

Which change causes the pH of 10 cm³ of 1.0 mol dm^-3 NaOH to be halved at 298 K?

K_w = 1.0 × 10^-14 at 298 K

• A: Adding 10 cm³ of water
• B: Adding 10 dm³ of water
• C: Adding 5 cm³ of 1.0 mol dm^-3 HCl
• D: Adding 10 cm³ of 1.0 mol dm^-3 HCl

The correct explanation is that adding 10 cm³ of 1.0 mol dm^-3 HCl neutralises half of the NaOH in the solution, reducing the pH significantly.

Step 1: Initial pH of NaOH

For a 1.0 mol dm^-3 NaOH solution:

[OH^-] = 1.0 mol dm^-3

[H⁺] = K_w / [OH^-] = 1.0 × 10^-14 / 1.0 = 1.0 × 10^-14 mol dm^-3

pH = -log(1.0 × 10^-14) = 14.0

Step 2: Effect of Adding 10 cm³ of 1.0 mol dm^-3 HCl

The NaOH solution contains 10 cm³ of 1.0 mol dm^-3 NaOH, which is:

Moles of NaOH = 1.0 × (10 / 1000) = 0.010 mol

Adding 10 cm³ of 1.0 mol dm^-3 HCl provides:

Moles of HCl = 1.0 × (10 / 1000) = 0.010 mol

The HCl neutralises all the NaOH, leaving no OH^- ions and creating a pH-neutral solution.

• A: Adding 10 cm³ of water dilutes the solution but does not halve the pH.
• B: Adding 10 dm³ of water dilutes the solution significantly but does not neutralise the NaOH.
• C: Adding 5 cm³ of HCl reduces the concentration of OH^- but does not halve the pH.

The correct answer is:

D: Adding 10 cm³ of 1.0 mol dm^-3 HCl